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A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the field is reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushes is 2V.
Correct answer is between '132,133'. Can you explain this answer?
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A separately excited DC generator, when running at 1200 rpm supplies 2...
Eg1 = VL + IaRa + 2 = 125 + (0.04 × 200) + 2 = 135 V
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A separately excited DC generator, when running at 1200 rpm supplies 2...
Given:
Speed at which DC generator is running, n1 = 1200 rpm
Speed at which DC generator is dropped, n2 = 1000 rpm
Voltage supplied, V = 125 V
Current supplied, I1 = 200 A
Armature resistance, Ra = 0.04 Ω
Field current, If1 = ?
We know that, the generated voltage (Eg) of a DC generator is given by the equation:
Eg = KΦnZ/60A
Where,
K = a constant depending on the design of the machine
Φ = flux per pole in webers
n = speed of the armature in rpm
Z = total number of armature conductors
A = number of parallel paths in the armature winding
Since the DC generator is separately excited, the flux per pole (Φ) is directly proportional to the field current (If). Therefore, we can write:
Φ1/Φ2 = If1/If2
Given that If1 = If and If2 = 0.8If
Therefore,
Φ1/Φ2 = If/0.8If = 1.25
Let us consider the circuit of the DC generator when running at 1200 rpm. We can write the following equations:
V = Eg + I1Ra
Eg = KΦ1n1Z/60A
Substituting the values, we get:
125 = KΦ11200Z/60A + (200)(0.04)
125 - 8 = KΦ11200Z/60A
117 = KΦ11200Z/60A
Similarly, for the new speed of 1000 rpm, we can write:
Eg = KΦ2n2Z/60A
V = Eg + I2Ra
Substituting the values, we get:
125 = KΦ21000Z/60A + I2(0.04)
125 - 8 = KΦ21000Z/60A + I2(0.04)
117 = KΦ21000Z/60A + I2(0.04)
Dividing the second equation by the first equation, we get:
Φ2/Φ1 = n2I2/(n1I1)
Substituting the values, we get:
Φ2/Φ1 = (1000)(I2)/(1200)(200)
Φ2/Φ1 = 5I2/6
Substituting the value of Φ1/Φ2 from earlier, we get:
1.25 = 5I2/6Φ1
Φ1 = 5I2/7.5
Substituting the value of Φ1 in the equation for KΦ1Z/60A, we get:
117 = K(5I2/7.5)(1200)Z/60A
KZ/2 = 9.36I2
Substituting the value of KZ/2 in the equation for Eg at 1000 rpm, we get:
Eg = (9.36I2)/Z
Substituting the values of Eg and Ra in the equation for V at 1000 rpm, we get:
125 = (9.36I2)/
Speed at which DC generator is running, n1 = 1200 rpm
Speed at which DC generator is dropped, n2 = 1000 rpm
Voltage supplied, V = 125 V
Current supplied, I1 = 200 A
Armature resistance, Ra = 0.04 Ω
Field current, If1 = ?
We know that, the generated voltage (Eg) of a DC generator is given by the equation:
Eg = KΦnZ/60A
Where,
K = a constant depending on the design of the machine
Φ = flux per pole in webers
n = speed of the armature in rpm
Z = total number of armature conductors
A = number of parallel paths in the armature winding
Since the DC generator is separately excited, the flux per pole (Φ) is directly proportional to the field current (If). Therefore, we can write:
Φ1/Φ2 = If1/If2
Given that If1 = If and If2 = 0.8If
Therefore,
Φ1/Φ2 = If/0.8If = 1.25
Let us consider the circuit of the DC generator when running at 1200 rpm. We can write the following equations:
V = Eg + I1Ra
Eg = KΦ1n1Z/60A
Substituting the values, we get:
125 = KΦ11200Z/60A + (200)(0.04)
125 - 8 = KΦ11200Z/60A
117 = KΦ11200Z/60A
Similarly, for the new speed of 1000 rpm, we can write:
Eg = KΦ2n2Z/60A
V = Eg + I2Ra
Substituting the values, we get:
125 = KΦ21000Z/60A + I2(0.04)
125 - 8 = KΦ21000Z/60A + I2(0.04)
117 = KΦ21000Z/60A + I2(0.04)
Dividing the second equation by the first equation, we get:
Φ2/Φ1 = n2I2/(n1I1)
Substituting the values, we get:
Φ2/Φ1 = (1000)(I2)/(1200)(200)
Φ2/Φ1 = 5I2/6
Substituting the value of Φ1/Φ2 from earlier, we get:
1.25 = 5I2/6Φ1
Φ1 = 5I2/7.5
Substituting the value of Φ1 in the equation for KΦ1Z/60A, we get:
117 = K(5I2/7.5)(1200)Z/60A
KZ/2 = 9.36I2
Substituting the value of KZ/2 in the equation for Eg at 1000 rpm, we get:
Eg = (9.36I2)/Z
Substituting the values of Eg and Ra in the equation for V at 1000 rpm, we get:
125 = (9.36I2)/
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A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer?
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A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer?.
A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A separately excited DC generator, when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistant what will be the current (in amp) when the speed is dropped to 1000 rpm and the fieldis reduced to 80%. Armature resistance is 0.04 Ω and total drop at brushesis 2V.Correct answer is between '132,133'. Can you explain this answer?.
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